Question

in an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. prove that 9 AD2 = 7 AB2

Answer 1

DE=BE -BD...............................(B-D-E)

$= \frac{1}{2}ab - \frac{1}{3} ab$
$de = \frac{1}{6} ab$

DE =1/6AB

NOW,
IN TRIANGLE ADE,
<AED=90°

BY PYTHAGORAS THEOREM,

AD^2=AE^2+DE^2
AD^2=(√3/2AB)^2+(1/6AB)^2
AD^2=7/9 AB^2
9AD^2=7AB^2


HENCE PROVED

HOPE THIS HELPS!!!

Answer 2

Answer:

Check your answer please